In each iteration, we move slow one step forward and fast two steps forward. In this code, we initialize slow and fast to the head of the list, and then we use a while loop to iterate through the list. If there is no cycle, then the fast pointer will reach the end of the list and we can safely assume that there is no cycle. The intuition behind this algorithm is that if there is a cycle in the linked list, then the fast pointer will eventually catch up to the slow pointer as they traverse the list. If fast ever reaches the same node as slow, then there is a cycle in the list and we can return true. If fast reaches the end of the list (i.e., fast.next is null), then there is no cycle in the list and we can return false. Move slow one step forward and fast two steps forward. Initialize two pointers, slow and fast, to the head of the linked list. To detect if a linked list has a cycle, we can use the "fast-slow" pointer approach. Your task is to implement a function hasCycle that takes the head of a linked list as input and returns a boolean value indicating whether the linked list contains a cycle. A cycle in a linked list occurs when a node in the list can be reached again by continuously following the next pointer. Given the head of a linked list, determine if the linked list has a cycle in it. In this post, we will discuss the problem statement and provide a solution using the "fast-slow" pointer approach.
Detecting a cycle in a linked list is a classic algorithmic problem that is frequently encountered in technical interviews.
0 kommentar(er)
